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Series DC Circuits Practice Worksheet with Answers
27 questions By Tony R. Kuphaldt
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Question 22 of 27
Complete the table of values for this circuit:
Reveal answer
Follow-up question #1: without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R1 were to fail open.
Follow-up question #2: without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R1 were to fail shorted.
Notes:Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
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Question 23 of 27
Complete the table of values for this circuit:
Reveal answer
Follow-up question #1: without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R2 were to fail open.
Follow-up question #2: without performing any mathematical calculations, determine the effects on all the component voltage drops and currents if resistor R2 were to fail shorted.
Notes:Discuss with your students what a good procedure might be for calculating the unknown values in this problem, and also how they might check their work.
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Question 24 of 27
In a series circuit, certain general rules may be stated with regard to quantities of voltage, current, resistance, and power. Express these rules, using your own words:
“In a series circuit, voltage . . .”
“In a series circuit, current . . .”
“In a series circuit, resistance . . .”
“In a series circuit, power . . .”
For each of these rules, explain why it is true.
Reveal answer“In a series circuit, voltage drops add to equal the total.”
“In a series circuit, current is equal through all components.”
“In a series circuit, resistances add to equal the total.”
“In a series circuit, power dissipations add to equal the total.”
Here’s a summary of series circuit rules.
Notes:Rules of series and parallel circuits are very important for students to comprehend. However, a trend I have noticed in many students is the habit of memorizing rather than understanding these rules. Students will work hard to memorize the rules without really comprehending why the rules are true, and therefore often fail to recall or apply the rules properly.
An illustrative technique I have found very useful is to have students create their own example circuits in which to test these rules. Simple series and parallel circuits pose little challenge to construct, and therefore serve as excellent learning tools. What could be better, or more authoritative, than learning principles of circuits from real experiments? This is known as primary research, and it constitutes the foundation of scientific inquiry. The greatest problem you will have as an instructor is encouraging your students to take the initiative to build these demonstration circuits on their own, because they are so used to having teachers simply tell them how things work. This is a shame, and it reflects poorly on the state of modern education.
llaboutcircuits.com/worksheets/series-dc-circuits/. In question 5, total series resistance is sum of each resistor, not product. Correct both question and answer
Hello Everybody and thanks for your fantastic articles!
I’m having troubles solving Excercise 19 of your Series circuits problems (https://www.allaboutcircuits.com/worksheets/series-dc-circuits/): For this excercise I’m getting a power dissipation for the heater of 403 W instead of 321.1 W.
Since I’m learning electronics I’m surely doing something wrong… Here how I’m reasoning:
1. With 110 V and a power dissipated of 500 W, I can calculate the current in the circuit, that should be I=P/V 500/110 = 4.545 A
2. Knowing that the current in the circuit is 4.545 A I can use Ohm’s law to calculate the resistance of the heater: R=V/I 110/4.545 = 24 Ohm
3. Now, longing the wires it’s the same as adding 6 Ohm of resistance, for a total resistance of 30 Ohm (24 + 6) so the current in the circuit will change accordingly. To find the new amount of current I use Ohm’s law once again: I=V/R 110/30 = 3.666 A
4. Having a new current value of 3.666 A, I can now calculate power dissipation P=V*I 110*3.666 = 403 W
Could you please help me understand where I’m wrong?
Thank you so much!
Stefano.